The Physics Philes, lesson 8: The Electric Slide

I’ve spent the last few weeks trying my best to explain Newton’s three laws of motion. It’s been a pretty wild ride, but now it’s time to move on. This week we’ll look at some forces that you and I experience one form or another on a daily basis: frictional forces.

I’m going to do something a bit different this week. Usually I explain a concept then do an example problem. But this week? Well, I don’t really understand the math yet. For some reason, it’s not penetrating my skull. So what I want to do is this: I’ll explain the concepts this week, work really,really hard this week to understand the math that backs it up, and do some example problems next week. For now, I’ll paint a word picture of three frictional forces: kinetic, static, and rolling.

Friction forces are contact forces. When you push a book across a table, we can think of the surface of the table as exerting a single contact force on the book. That single contact force has a component that is parallel to the table surface and a component that is perpendicular to the surface of the table. The perpendicular force is the normal force (n), which we’ve worked with before. The parallel force is the friction force (f). The friction force always opposes the relative motion of the two surfaces involved.

There are a few different types of friction forces. The first I’ll discuss is the kinetic friction force. A kinetic friction force occurs when a body slides across a surface. Remember the book that we pushed across the table? The friction force action on that book while it’s in motion is the kinetic friction force. The magnitude of this force doesn’t depend on how hard the book was pushed. What matters is the normal force. The magnitude of the kinetic friction force usually increases when the normal force increases. This makes sense. It’s easier to push one of my Incredible Hulk trade paperbacks across a table than it is to push my gigantic legal dictionary. The magnitude of the normal force acting on the dictionary is much greater than the magnitude of the normal force acting on the trade paperback.

For once, a physics concept makes sense then I think about it in my life! What a great feeling!

It’s been shown by experiment that, in many cases, the magnitude of the kinetic friction force is approximately proportional to the magnitude of the normal force. This relationship is called the coefficient of kinetic friction. However, the relationship is only an approximation. Kinetic friction isn’t perfectly constant because the friction and normal forces are the results of forces between molecules as two rough surfaces come into contact with each other. So we can’t get an exact number, but we can get a pretty good approximation.

At this point, you may be wondering, What about when an object just won’t budge? Surely friction is at work in that situation, as well. To which I will say, How astute of you. Yes, there is a friction force at work in that situation. And that force is called the static friction force.

Let’s think about a box with a rope attached sitting on a table. The box is at rest and in equilibrium under the force of its weight and the normal force exerted by the table. Now imagine you pick up that rope and start to pull the box across the surface of the table. At first, the box doesn’t move. The static friction force grows to match the tension force exerted by the rope. But at some point, the static friction force will max out. It won’t be able to increase any more. When that point is reached, the box will break loose and move in the direction of the tension force. At this critical point, the bonds formed between the molecules on the surface of the box and the molecules on the surface of the table will break, allowing the box to move. The friction force being applied to the box before the box actually moves is the static friction force.

Like kinetic friction force, experiment has shown that in many instances the maximum static friction force is approximately proportional to the normal force. We call this relationship the coefficient of static friction. In any particular situation, the actual static friction force can be zero – meaning no other force is being applied to the body – or the maximum value. So before we started pulling the box, the static friction force would be zero because there was no other force acting on the box. But when we started pulling, in that moment before the box started moving the static friction force would be at its maximum.

The static friction force will be greater than the kinetic friction force because it’s easier to keep an object moving than to get that object moving in the first place. This means that the coefficient of kinetic friction will usually be smaller than the coefficient of static friction for any given pair of surfaces.

The last friction force I’ll discuss today is rolling friction. This is probably the most intuitive of these three friction forces. It basically explains why it’s so much easier to roll something than it is to push or pull something along a flat surface. Like the others, we can calculate a coefficient of rolling friction. We do this by dividing the horizontal force needed for constant speed on a flat surface by the upward normal force exerted by the surface. The coefficient of rolling friction is usually much smaller than the coefficient of kinetic friction.

The concepts, I think, are not too complicated. However, I’m still struggling to understand the math that makes it possible. But I’m working on it very hard and I promise that next week I’ll be back with some example problems so we can see exactly how we can put our new knowledge to use.

In the meantime, I’m counting on you to let me know if there is any incomplete or just plain wrong information in this post.

Actually, I think I’ve figured it out. Silly algebra rules that I haven’t used in 10 years. I am still having trouble with the math for fluid resistance, though. It looks like calculus, which would explain why I’m so lost.

There are two terms in fluid resistance. One that depends on velocity squared and one that depends on velocity linearly: F=-(av^2+bv)

Coefficient (a) depends on radius of object squared and pressure, while coefficient (b) depends on radius linearly and viscosity, at least for spherical objects.
In most cases, velocity will be low enough that only the linear part happens, mostly. And so:

F=-bv.

The fluid resistance depends on the velocity.
Now begins the analysis of falling things.

I will label b/m as c, and rate of change of velocity (acceleration) as v*, and so dividing both sides by m:

=c-g

Now, let’s explore this equation. What happens when the acceleration is zero as time goes to infinity? Then:

c=g
=g/c

That is called the terminal velocity. It is the maximum velocity allowed by the fluid for an object of certain shape. It has to be, because acceleration is zero, and because of that, it can’t go any faster. Now, I say time goes to infinity because as air drag becomes nearly equal to gravity, the object’s velocity won’t increase as much. But since drag depends on velocity, that means it will only increase even smaller. This is the real world, though, and in the real world, we do reach that velocity in finite time (never confuse the mathematical world with the real one, always check in what domain is an equation useful in!). Now, if velocity is zero, then:

Makes sense. At the beginning, the object is standing still and it feels the gravitational force from the Earth. From this, let’s build an equation for the velocity. From v=g/c, it has to be that:

v(t)=(g/c)-f(t) (f(t) being an unknown kind of fuction)

At t=0, v=0, so it has to be that f(0)=g/c, so that way: v(0)=(g/c)-(g/c)=0. So, updating the equation, it is:

v(t)=(g/c)-(g/c)f(t)=(g/c)(1-f(t))

Also, at time=infinity:

v(infinity)=g/c, meaning f(infinity)=0

The only equation that fits f(t) is an exponential function:

f(t)=e^(-kt), where if time equals zero, e^0=1, and if time goes to infinity, e^(-infinity)=0. Try plugging in in a calculator e^(negative any huge number), and you will see it is very close to zero. From now on, I say that exponential function is like this: e^(-kt)=e[-kt] for simplicity. And so, plugging it back in, you get:

v(t)=(g/c)(1-e[-kt])

Now, what is k?

Okay, I don’t know how much calculus you know at this point. But I can tell you by using something called the chain rule, the fact that the rate of change of e[x] is e[x], and the rate of change of a number (in this case g/c) is zero, the acceleration is:

a(t)=(k*g/c)e[-kt]

k is outside too because of the chain rule. The rate of change of -kt is the slope, which is -k, and the negative was cancelled by the minus into a positive.

At time zero, we know acceleration is only the one by gravity, and so:

g=(g*k/c)e[0]=(g*k/c), and this gets me k=c.

Yay, now you can calculate the velocity of an object going through a fluid!

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